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Date November 2021 Marks available 2 Reference code 21N.1.AHL.TZ0.17
Level Additional Higher Level Paper Paper 1 Time zone Time zone 0
Command term Hence and Find Question number 17 Adapted from N/A

Question

The sides of a bowl are formed by rotating the curve y=6lnx, 0y9, about the y-axis, where x and y are measured in centimetres. The bowl contains water to a height of hcm.

Show that the volume of water, V, in terms of h is V=3πeh3-1.

[5]
a.

Hence find the maximum capacity of the bowl in cm3.

[2]
b.

Markscheme

attempt to use V=πabx2dy             (M1)

x=ey6 or any reasonable attempt to find x in terms of y             (M1)

V=π0hey3dy             A1


Note: Correct limits must be seen for the A1 to be awarded.


=π3ey30h             (A1)


Note: Condone the absence of limits for this A1 mark.


=3πeh3-e0             A1

=3πeh3-1             AG


Note: If the variable used in the integral is x instead of y (i.e. V=π0hex3dx) and the candidate has not stated that they are interchanging x and y then award at most M1M1A0A1A1AG.

 

[5 marks]

a.

maximum volume when h=9cm             (M1)

max volume =180cm3             A1

 

[2 marks]

b.

Examiners report

A number of candidates switched variables so that y=6lnx and then used πy2dx. Other candidates who correctly found x in terms of y failed to use the limits 0 and h, using 0 and 9 instead. As part (a) was to show that the volume was equal to the final expression it was necessary for examiners to see steps in obtaining the result. It was common to miss out any expression involving e0. Since the value 1 could be written from the answer given, where this value came from needed to be shown. It was encouraging to see correct answers to (b), even when candidates had failed to gain marks for (a). Some candidates successfully used their GDC to calculate the value of the definite integral numerically.

a.

A number of candidates switched variables so that y=6lnx and then used πy2dx. Other candidates who correctly found x in terms of y failed to use the limits 0 and h, using 0 and 9 instead. As part (a) was to show that the volume was equal to the final expression it was necessary for examiners to see steps in obtaining the result. It was common to miss out any expression involving e0. Since the value 1 could be written from the answer given, where this value came from needed to be shown. It was encouraging to see correct answers to (b), even when candidates had failed to gain marks for (a). Some candidates successfully used their GDC to calculate the value of the definite integral numerically.

b.

Syllabus sections

Topic 5—Calculus » AHL 5.12—Areas under a curve onto x or y axis. Volumes of revolution about x and y
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Topic 5—Calculus

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